Question

The moment of inertia of thin disc of mass $m$ and radius $r$, about an axis passing through one of its diameter is given by

Answer 1

Step 1: Given data

1. Mass of disc = $m$
2. Radius of disc = $r$
3. We know, for a disc, moment of inertia through the center of mass and perpendicular to the disc is $I_{CM}=\frac{m{r}^2}{2}$
4. We have to find the moment of inertia of the disc passing through one of its diameter.

Step 2: Theorem used

Perpendicular axis theorem- Moment of inertia of a body about an axis perpendicular to plane is equal to sum of moment of inertia about any two perpendicular axes in the plane of the body.

$I_Z=I_X+I_Y$

where $I_Z$, $I_X$ and $I_Y$ are the three perpendicular axes.

Step 3: Calculation

Here, by perpendicular axis theorem, we can say that,

Moment of inertia about an axis passing through center of mass and perpendicular to the plane = moment of inertia about any two perpendicular axis passing through the diameter

$\therefore I_{CM}=2I_D$ (Since moment of inertia through the diametrical axes are equal)

where $I_{CM}$ is moment of inertia through center of mass and $I_D$ is moment of inertia through diameter

$$\begin{gathered} \because I_D=\frac{I_{CM}}{2} \\ \Rightarrow I_D=\frac{m{r}^2}{4} \end{gathered}$$

The moment of inertia of thin disc of mass $m$ and radius $r$, about an axis passing through one of its diameter is given by $\frac{m{r}^2}{4}$.